Question

Solve the differential equation \[{{\cos }^{2}}x\dfrac{dy}{dx}+y=\tan x\]
A). \[y=t-1+c{{e}^{-1}}\] where \[t=\tan x\]
B). \[y=t+1+c{{e}^{-1}}\] where \[t=\tan x\]
C). \[y=t-1-c{{e}^{-1}}\] where \[t=\tan x\]
D). None of these

Answer 1

Hint: In this question make the equation in the form of a differential equation and then find the value of \[P\] and \[Q\] . To find these values first of all find out the integrating factor and then find out the solution. Apply integration by parts to find out the value in terms of \[y\] and then check which option is correct among them.

Complete step-by-step solution:
By definition, a differential equation is an equation that contains one or more functions and their derivatives. The derivatives of a function define the rate of change of the function at a given position. Differential equations are commonly utilised in biology, physics, engineering, and other sciences. The differential equation's major goal is to investigate the solutions that satisfy the equations as well as the properties of the solutions.
A differential function is an equation that contains a function's derivative. A differential equation is defined as an equation involving the derivative (derivatives) of the dependent variable with regard to the independent variable (variables) in calculus. Partially derivatives and regular derivatives can both be found in a differential equation. The differential equation helps us illustrate a relationship between the changing quantity and the change in another quantity, and the derivative represents nothing more than a rate of change.
Now according to the question we have given the term:
\[\Rightarrow {{\cos }^{2}}x\dfrac{dy}{dx}+y=\tan x\]
Divide both side by \[{{\cos }^{2}}x\] to simplify the term:
\[\Rightarrow \dfrac{{{\cos }^{2}}x}{{{\cos }^{2}}x}\dfrac{dy}{dx}+\dfrac{y}{{{\cos }^{2}}x}=\dfrac{\tan x}{{{\cos }^{2}}x}\]
\[\Rightarrow \dfrac{dy}{dx}+y\cdot \dfrac{1}{{{\cos }^{2}}x}=\dfrac{\tan x}{{{\cos }^{2}}x}\]
As we know that \[\dfrac{1}{{{\cos }^{2}}x}={{\sec }^{2}}x\]
\[\Rightarrow \dfrac{dy}{dx}+{{\sec }^{2}}x\cdot y=\tan x\cdot {{\sec }^{2}}x\]
Hence it is in the form of \[\dfrac{dy}{dx}+P\cdot y=Q\]
Where \[P={{\sec }^{2}}x\] and \[Q=\tan x\cdot {{\sec }^{2}}x\]
Therefore the integrating factor will be given by:
\[\Rightarrow I.F={{e}^{\int{Pdx}}}\]
Now we will find out the value of \[P\] and \[Q\]
To find the value of \[P\] :
\[\Rightarrow I.F={{e}^{\int{Pdx}}}\]
\[\Rightarrow I.F={{e}^{\int{{{\sec }^{2}}xdx}}}\]
\[\Rightarrow I.F={{e}^{\tan x}}\]
 The solution of the equation is given by:
\[\Rightarrow y\times I.F=\int{Q\times \left( I.F \right)dx+c}\]
\[\Rightarrow y\times {{e}^{\tan x}}=\int{\tan x\cdot {{\sec }^{2}}x\times {{e}^{\tan x}}dx+c}\]
Let us assume that:
\[\tan x=t\]
On differentiating it on both sides we get
 \[\Rightarrow \dfrac{d}{dx}\tan x=\dfrac{d}{dx}t\]
\[\Rightarrow {{\sec }^{2}}x=\dfrac{dt}{dx}\]
\[\Rightarrow {{\sec }^{2}}xdx=dt\]
Now put the values that we have assumed:
\[\Rightarrow y{{e}^{\operatorname{t}}}=\int{t\cdot {{e}^{\operatorname{t}}}dt+c}\]
Let \[I=\int{t\cdot {{e}^{\operatorname{t}}}dt}\]
To integrate the given term \[I\] apply integration by parts that is:
\[\Rightarrow \int{f(t)g(t)dt=f(t)}\int{g(t)dx-\int{\left( \dfrac{d}{dt}f(t)\int{g(t)dx} \right)}}dx\]
Where \[f(t)=t\] and \[g(t)={{e}^{\operatorname{t}}}\]
\[\Rightarrow I=t\int{{{e}^{\operatorname{t}}}dt-\int{\left( \dfrac{d}{dt}(t)\int{{{e}^{\operatorname{t}}}dt} \right)dt}}\]
\[\Rightarrow I=t{{e}^{\operatorname{t}}}-\int{\left( \int{{{e}^{\operatorname{t}}}dt} \right)}dt\]
\[\Rightarrow I=t{{e}^{\operatorname{t}}}-\int{{{e}^{\operatorname{t}}}dt}\]
\[\Rightarrow I=t{{e}^{\operatorname{t}}}-{{e}^{\operatorname{t}}}\]
Taking \[{{e}^{\operatorname{t}}}\] in common we get:
\[\Rightarrow I={{e}^{t}}(t-1)\]
Therefore the solution will be:
\[\Rightarrow y{{e}^{\operatorname{t}}}={{e}^{\operatorname{t}}}\left( \operatorname{t}-1 \right)+c\]
Divide the whole term by \[{{e}^{t}}\] we get:
\[\Rightarrow y\dfrac{{{e}^{\operatorname{t}}}}{{{e}^{\operatorname{t}}}}=\dfrac{{{e}^{\operatorname{t}}}\left( \operatorname{t}-1 \right)}{{{e}^{\operatorname{t}}}}+\dfrac{c}{{{e}^{\operatorname{t}}}}\]
\[\Rightarrow y=\left( \operatorname{t}-1 \right)+\dfrac{c}{{{e}^{\operatorname{t}}}}\]
\[\Rightarrow y=\left( \operatorname{t}-1 \right)+c{{e}^{-\operatorname{t}}}\]
Hence option \[(A)\] is correct as we have got the value \[y=t-1+c{{e}^{-1}}\] where \[t=\tan x\].

Note: You should know that in real life, ordinary differential equations are used to compute the movement or flow of electricity, the motion of an object to and fro, such as a pendulum, and to illustrate thermodynamics ideas. In medical terminology, they're also used to visualise the progression of diseases in graphical form.